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<h1>Converting UTM to Latitude and Longitude (Or Vice Versa)</h1>

Steven Dutch, Natural and Applied Sciences, <a 
href="http://weba.uwgb.edu/">University
of Wisconsin - Green Bay</a><br>
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<p><a href="http://www.uwgb.edu/dutchs/FieldMethods/UTMSystem.htm">Information
 on the UTM system</a><br>
<a href="http://www.uwgb.edu/dutchs/usefuldata/UTMConversions1.xls">Spreadsheet
 For UTM Conversion</a><br>
<a href="http://www.uwgb.edu/dutchs/usefuldata/OddGrids.HTM">Help! My 
Data Doesn't Look Like A UTM Grid!</a></p>
<p>I get enough inquiries on this subject that I decided to create a 
page for
it.</p>
<p>Caution! Unlike latitude and longitude, there is no physical frame of
reference for UTM grids. Latitude is determined by the earth's polar 
axis.
Longitude is determined by the earth's rotation. If you can see the 
stars and
have a sextant and a good clock set to Greenwich time, you can find your
latitude and longitude. But there is no way to determine your UTM 
coordinates
except by calculation.</p>
<p>UTM grids, on the other hand, are created by laying a square grid on 
the
earth. This means that different maps will have different grids 
depending on the
datum used (model of the shape of the earth). I saw US military maps of 
Germany
shift their UTM grids by about 300 meters when a more modern datum was 
used for
the maps. Also, old World War II era maps of Europe apparently used a 
single grid
for all of Europe and grids in some areas are wildly tilted with respect
 to
latitude and longitude.</p>
<p>The two basic references for converting UTM and geographic 
coordinates are
U.S. Geological
Survey Professional Paper 1395 and U. S. Army
Technical Manual TM 5-241-8 (complete citations below). Each has 
advantages and
disadvantages.&nbsp;</p>
<p>For converting latitude and longitude to UTM, the Army publication is
 better. Its notation is more
consistent and the formulas more clearly laid out, making code easier to
 debug.
In defense of the USGS, their notation is constrained by space, and the 
need to
be consistent with cartographic literature and descriptions of several 
dozen
other map projections in the book.</p>
<p>For converting UTM to latitude and longitude, however, the Army 
publication
has formulas that involve latitude (the quantity to be found) and which 
require
reverse interpolation from tables. Here the USGS publication is the only
 game in
town.</p>
<p>Some extremely tiny terms that will not seriously affect meter-scale 
accuracy
have been omitted.</p>
<h2>Converting Between Decimal Degrees, Degrees, Minutes and Seconds, 
and
Radians</h2>
<p><b>(dd + mm/60 +ss/3600) to Decimal degrees (dd.ff)</b></p>
<p>dd = whole degrees, mm = minutes, ss = seconds</p>
<p>dd.ff = dd + mm/60 + ss/3600</p>
<p>Example: 30 degrees 15 minutes 22 seconds = 30 + 15/60 + 22/3600 = 
30.2561</p>
<p><b>Decimal degrees (dd.ff) to (dd + mm/60 +ss/3600)</b></p>
<p>For the reverse conversion, we want to convert dd.ff to dd mm ss. 
Here ff =
the fractional part of a decimal degree.</p>
<p>mm = 60*ff</p>
<p>ss = 60*(fractional part of mm)</p>
<p>Use only the whole number part of mm in the final result.</p>
<p>30.2561 degrees = 30 degrees</p>
<p>.2561*60 = 15.366 minutes</p>
<p>.366 minutes = 22 seconds, so the final result is 30 degrees 15 
minutes 22
seconds</p>
<p><b>Decimal degrees (dd.ff) to Radians</b></p>
<p>Radians = (dd.ff)*pi/180</p>
<p><b>Radians to Decimal degrees (dd.ff)</b></p>
<p>(dd.ff) = Radians*180/pi</p>
<p><b>Degrees, Minutes and Seconds to Distance</b></p>
<p>A degree of longitude at the equator is 111.2 kilometers. A minute is
 1853
meters. A second is 30.9 meters. For other latitudes multiply by 
cos(lat).
Distances for degrees, minutes and seconds<b> </b>in latitude are very 
similar
and differ very slightly with latitude. (Before satellites, observing 
those
differences was a principal method for determining the exact shape of 
the
earth.)</p>
<h2>Converting Latitude and Longitude to UTM</h2>
<p>Okay, take a deep breath. This will get <i>very</i>  complicated, but
 the
math, although tedious, is only algebra and trigonometry.&nbsp;It would 
sure be
nice if someone wrote a <a href="#Spreadsheet"> spreadsheet</a> to do 
this.</p>
<table border="1" width="100%">
  <tbody><tr>
    <td><img src="utmformulas_files/UTMConvert.gif" border="0" 
width="300" height="300"></td>
    <td>P = point under consideration<br>
      F = foot of perpendicular from P to the central meridian. The 
latitude of
      F is called the <i>footprint latitude</i>.<br>
      O = origin (on equator)<br>
      OZ = central meridian<br>
      LP = parallel of latitude of P<br>
      ZP = meridian of P<br>
      OL = k<sub>0</sub>S = meridional arc from equator<br>
      LF = ordinate of curvature<br>
      OF = N = grid northing<br>
      FP = E = grid distance from central meridian<br>
      GN = grid north<br>
      C = convergence of meridians = angle between true and grid north</td>
  </tr>
</tbody></table>
<p>Another thing you need to know is the datum being used:</p>
<table border="1" width="100%">
  <tbody><tr>
    <td width="20%"><b>Datum</b></td>
    <td width="20%"><b>Equatorial Radius, meters (a)</b></td>
    <td width="20%"><b>Polar Radius, meters (b)</b></td>
    <td width="20%"><b>Flattening (a-b)/a</b></td>
    <td width="20%"><b>Use</b></td>
  </tr>
  <tr>
    <td width="20%">NAD83/WGS84</td>
    <td width="20%">6,378,137</td>
    <td width="20%">6,356,752.3142</td>
    <td width="20%">1/298.257223563</td>
    <td width="20%">Global</td>
  </tr>
  <tr>
    <td width="20%">GRS 80</td>
    <td width="20%">6,378,137</td>
    <td width="20%">6,356,752.3141</td>
    <td width="20%">1/298.257222101</td>
    <td width="20%">US</td>
  </tr>
  <tr>
    <td width="20%">WGS72</td>
    <td width="20%">6,378,135</td>
    <td width="20%">6,356,750.5</td>
    <td width="20%">1/298.26</td>
    <td width="20%">NASA, DOD</td>
  </tr>
  <tr>
    <td width="20%">Australian 1965</td>
    <td width="20%">6,378,160</td>
    <td width="20%">6,356,774.7</td>
    <td width="20%">1/298.25</td>
    <td width="20%">Australia</td>
  </tr>
  <tr>
    <td width="20%">Krasovsky 1940</td>
    <td width="20%">6,378,245</td>
    <td width="20%">6,356,863.0</td>
    <td width="20%">1/298.3</td>
    <td width="20%">Soviet Union</td>
  </tr>
  <tr>
    <td width="20%">International (1924) -Hayford (1909)</td>
    <td width="20%">6,378,388</td>
    <td width="20%">6,356,911.9</td>
    <td width="20%">1/297</td>
    <td width="20%">Global except as listed</td>
  </tr>
  <tr>
    <td width="20%">Clake 1880</td>
    <td width="20%">6,378,249.1</td>
    <td width="20%">6,356,514.9</td>
    <td width="20%">1/293.46</td>
    <td width="20%">France, Africa</td>
  </tr>
  <tr>
    <td width="20%">Clarke 1866</td>
    <td width="20%">6,378,206.4</td>
    <td width="20%">6,356,583.8</td>
    <td width="20%">1/294.98</td>
    <td width="20%">North America</td>
  </tr>
  <tr>
    <td width="20%">Airy 1830</td>
    <td width="20%">6,377,563.4</td>
    <td width="20%">6,356,256.9</td>
    <td width="20%">1/299.32</td>
    <td width="20%">Great Britain</td>
  </tr>
  <tr>
    <td width="20%">Bessel 1841</td>
    <td width="20%">6,377,397.2</td>
    <td width="20%">6,356,079.0</td>
    <td width="20%">1/299.15</td>
    <td width="20%">Central Europe, Chile, Indonesia</td>
  </tr>
  <tr>
    <td width="20%">Everest 1830</td>
    <td width="20%">6,377,276.3</td>
    <td width="20%">6,356,075.4</td>
    <td width="20%">1/300.80</td>
    <td width="20%">South Asia</td>
  </tr>
</tbody></table>
<p>Don't interpret the chart to mean there is radical disagreement about
 the
shape of the earth. The earth isn't perfectly round, it isn't even a 
perfect 
ellipsoid, and slightly different
shapes work better for some regions than for the earth as a whole. The 
top three
are based on worldwide data and are truly global. Also, you are very 
unlikely to
find UTM grids based on any of the earlier projections.</p>
<p>The most modern datums (jars my Latinist sensibilities since the 
plural of <i>datum</i>
in Latin is <i>data</i>, but that has a different meaning to us) are 
NAD83 and
WGS84. These are based in turn on GRS80. Differences between the three 
systems
derive mostly from redetermination of station locations rather than 
differences 
in the datum. Unless you are locating a first-order station to 
sub-millimeter 
accuracy (in which case you are way beyond the scope of this page) you 
can 
probably regard them as essentially identical.</p>
<p><b>I have no information on the NAD83 and WGS84 datums, nor can my 
spreadsheet calculate differences between those datums and WGS84.</b></p>
<h2>Formulas For Converting Latitude and Longitude to UTM</h2>
<p>These formulas are slightly modified from Army (1973). They are 
accurate to within
less than a meter within a
given grid zone. The original formulas include a now obsolete term that 
can be handled more 
simply - it merely converts radians to seconds of arc. That term is 
omitted here 
but discussed below.</p>
<h3>Symbols</h3>
<ul>
  <li>lat = latitude of point</li>
  <li>long = longitude of point</li>
  <li>long<sub>0</sub> = central meridian of zone</li>
  <li>k<sub>0</sub>&nbsp; = scale along long<sub>0</sub>&nbsp;= 0.9996. 
Even 
	though it's a constant, we retain it as a separate symbol to keep the 
	numerical coefficients simpler, also to allow for systems that might 
use a 
	different Mercator projection.</li>
  <li>e = SQRT(1-b<sup>2</sup>/a<sup>2</sup>) = .08 approximately. This 
is the
    eccentricity of the earth's elliptical cross-section.</li>
  <li>e'<sup>2</sup> = (ea/b)<sup>2</sup> = e<sup>2</sup>/(1-e<sup>2</sup>)
 =
    .007 approximately. The quantity e' only occurs in even powers so it
 need
    only be calculated as e'<sup>2</sup>.</li>
  <li>n = (a-b)/(a+b)</li>
  <li>rho = a(1-e<sup>2</sup>)/(1-e<sup>2</sup>sin<sup>2</sup>(lat))<sup>3/2</sup>.
    This is the radius of curvature of the earth in the meridian plane.</li>
  <li>nu = a/(1-e<sup>2</sup>sin<sup>2</sup>(lat))<sup>1/2</sup>. This 
is the
    radius of curvature of the earth perpendicular to the meridian 
plane. It is
    also the distance from the point in question to the polar axis, 
measured
    perpendicular to the earth's surface.</li>
  <li>p = (long-long<sub>0</sub>) <b>in radians </b>(This differs from 
the 
	treatment in the Army reference)</li>
</ul>
<h3>Calculate the Meridional Arc</h3>
<p>S is the meridional arc through the point in question (the distance 
along the
earth's surface from the equator). All angles are in radians.</p>
<ul>
  <li>S = A'lat - B'sin(2lat) + C'sin(4lat) - D'sin(6lat) + E'sin(8lat),
 where
    lat is in radians and&nbsp;</li>
  <li>A' = a[1 - n + (5/4)(n<sup>2</sup> - n<sup>3</sup>) + (81/64)(n<sup>4</sup>
    - n<sup>5</sup>) ...]</li>
  <li>B' = (3 tan/2)[1 - n + (7/8)(n<sup>2</sup> - n<sup>3</sup>) + 
(55/64)(n<sup>4</sup>
    - n<sup>5</sup>) ...]</li>
  <li>C' = (15 tan<sup>2</sup>/16)[1 - n + (3/4)(n<sup>2</sup> - n<sup>3</sup>)&nbsp;

    ...]</li>
  <li>D' = (35 tan<sup>3</sup>/48)[1 - n + (11/16)(n<sup>2</sup> - n<sup>3</sup>)&nbsp;

    ...]</li>
  <li>E' = (315 tan<sup>4</sup>/512)[1 - n&nbsp; ...]</li>
</ul>
<p>The USGS gives this form, which may be more appealing to some. (They 
use M
where the Army uses S)</p>
<ul>
  <li>M = a[(1 - e<sup>2</sup>/4 - 3e<sup>4</sup>/64 - 5e<sup>6</sup>/256

    ....)lat<br>
    &nbsp;- (3e<sup>2</sup>/8 + 3e<sup>4</sup>/32 + 45e<sup>6</sup>/1024...)sin(2lat)&nbsp;<br>
    + (15e<sup>4</sup>/256 + 45e<sup>6</sup>/1024 + ....)sin(4lat)<br>
    &nbsp;- (35e<sup>6</sup>/3072 + ....) sin(6lat) + ....)] where lat 
is in
    radians</li>
</ul>
<p>This is the hard part. Calculating the arc length of an ellipse 
involves 
functions called <i>elliptic integrals</i>, which don't reduce to neat 
closed 
formulas. So they have to be represented as series.</p>
<h3> Converting Latitude and Longitude to UTM</h3>
<p>All angles are in radians.</p>
<p>y = northing = K1 + K2p<sup>2</sup> + K3p<sup>4</sup>, where</p>
<ul>
  <li>K1 = Sk<sub>0</sub>,</li>
  <li>K2 = k<sub>0</sub> nu sin(lat)cos(lat)/2 = k<sub>0</sub> nu sin(2 
lat)/4</li>
  <li>K3 = [k<sub>0</sub> nu sin(lat)cos<sup>3</sup>(lat)/24][(5
    - tan<sup>2</sup>(lat) + 9e'<sup>2</sup>cos<sup>2</sup>(lat) + 4e'<sup>4</sup>cos<sup>4</sup>(lat)]</li>
</ul>
<p>x = easting = K4p + K5p<sup>3</sup>, where</p>
<ul>
  <li>K4 = k<sub>0</sub> nu cos(lat)</li>
  <li>K5 = (k<sub>0</sub> nu cos<sup>3</sup>(lat)/6)[1 -
    tan<sup>2</sup>(lat) + e'<sup>2</sup>cos<sup>2</sup>(lat)]</li>
</ul>
<p>Easting x is relative to the central meridian. For conventional UTM 
easting add
500,000 meters to x.</p>
<h3>What the Formulas Mean</h3>
<p>The hard part, allowing for the oblateness of the Earth, is taken 
care of in 
calculating S (or M). So K1 is simply the arc length along the central 
meridian 
of the zone corrected by the scale factor. Remember, the scale is a hair
 less 
than 1 in the middle of the zone, and a hair more on the outside.</p>
<p>All the higher K terms involve nu, the local radius of curvature 
(roughly 
equal to the radius of the earth or roughly 6,400,000 m), trig 
functions, and powers of e'<sup>2</sup> ( = .007 ). So basically they 
are never 
much larger than nu. Actually the maximum value of K2 is about nu/4 
(1,600,000), K3 
is about nu/24 
(267,000) and K5 is about nu/6 (1,070,000). Expanding the expressions 
will show 
that the tangent terms don't affect anything.</p>
<p>If we were just to stop with the K2 term in the northing, we'd have a
 
quadratic in p. In other words, we'd approximate the parallel of 
latitude as a 
parabola. The real curve is more complex. It will be more like a 
hyperbola equatorward of about 45 degrees and an ellipse poleward, at 
least within the 
narrow confines of a UTM zone. (At any given latitude we're cutting the 
cone of 
latitude vectors with an inclined plane, so the resulting intersection 
will be a 
conic section. Since the projection cylinder has a curvature, the exact 
curve is 
not a conic but the difference across a six-degree UTM zone is pretty 
small.)&nbsp; Hence the need for higher order terms. Now p will 
never be more than +/-3 degrees = .05 radians, so p<sup>2</sup> is 
always less 
than .0025 (1/400) and p<sup>4</sup> is always less than .00000625 
(1/160000). 
Using a spreadsheet, it's easy to see how the individual terms vary with
 
latitude. K2p<sup>2</sup> never exceeds 4400 and K3p<sup>4</sup> is at 
most a 
bit over 3. That 
is, the curvature of a parallel of latitude across a UTM zone is at most
 a 
little less than 4.5 km 
and the maximum departure from a parabola is at most a few meters.</p>
<p>K4 is what we'd calculate for easting in a simple-minded way, just by
 
calculating arc distance along the parallel of latutude. But, as we get 
farther 
from the central meridian, the meridians curve inward, so our actual 
easting 
will be less than K4. That's what K5 does. Since p is never more than 
+/-3 
degrees = .05 radians, p<sup>3</sup> is always less than .000125 
(1/8000). The 
maximum value of K5p<sup>3</sup> is about 150 meters. </p>
<h3>That Weird Sin 1" Term in the Original Army Reference</h3>
<p>The Army reference defines p in seconds of arc and includes a sin 1" 
term in 
the K formulas. The Sin 1" term is a holdover from the days when this 
all had to 
be done on mechanical desk calculators (pre-computer) and terms had to 
be kept 
in a range that would&nbsp; retain sufficient precision at intermediate 
steps. 
For that small an angle the difference between sin 1" and 1" in radians 
is 
negligible. If p is in seconds of arc, then (psin 1") merely converts it
 to 
radians. </p>
<p>The sin 1" term actually included an extra factor of 10,000, which 
was then 
corrected by multiplying by large powers of ten afterward.</p>
<p>The logic is a bit baffling. If I were doing this on a desk 
calculator, I'd 
factor out as many terms as possible rather than recalculate them for 
each term. But perhaps in practice the algebraically 
obvious way created overflows or underflows, since calculators could 
only handle 
limited ranges. </p>
<p>In any case, the sin1" term is not needed any more. Calculate p in 
radians 
and omit the sin1" terms and the large power of ten multipliers.</p>
<h2>Converting UTM to Latitude and Longitude</h2>
<p>y = northing, x = easting (relative to central meridian; subtract 
500,000
from conventional UTM coordinate).</p>
<h3>Calculate the Meridional Arc</h3>
<p>This is easy: M = y/k<sub>0</sub>.</p>
<h3>Calculate Footprint Latitude</h3>
<ul>
  <li>mu = M/[a(1 - e<sup>2</sup>/4 - 3e<sup>4</sup>/64 - 5e<sup>6</sup>/256...)</li>
  <li>e<sub>1</sub> = [1 - (1 - e<sup>2</sup>)<sup>1/2</sup>]/[1 + (1 - e<sup>2</sup>)<sup>1/2</sup>]</li>
</ul>
<p>footprint latitude fp =&nbsp;mu + J1sin(2mu) + J2sin(4mu) + 
J3sin(6mu) +
J4sin(8mu), where:</p>
<ul>
  <li>J1 = (3e<sub>1</sub>/2 - 27e<sub>1</sub><sup>3</sup>/32 ..)</li>
  <li>J2 = (21e<sub>1</sub><sup>2</sup>/16 - 55e<sub>1</sub><sup>4</sup>/32
 ..)</li>
  <li>J3 = (151e<sub>1</sub><sup>3</sup>/96 ..)</li>
  <li>J4 = (1097e<sub>1</sub><sup>4</sup>/512 ..)</li>
</ul>
<h3>Calculate Latitude and Longitude</h3>
<ul>
  <li>e'<sup>2</sup> = (ea/b)<sup>2</sup> = e<sup>2</sup>/(1-e<sup>2</sup>)&nbsp;</li>
  <li>C1 = e'<sup>2</sup>cos<sup>2</sup>(fp)</li>
  <li>T1 = tan<sup>2</sup>(fp)</li>
  <li>R1 = a(1-e<sup>2</sup>)/(1-e<sup>2</sup>sin<sup>2</sup>(fp))<sup>3/2</sup>.
    This is the same as rho in the forward conversion formulas above, 
but
    calculated for fp instead of lat.</li>
  <li>N1 = a/(1-e<sup>2</sup>sin<sup>2</sup>(fp))<sup>1/2</sup>.
    This is the same as nu in the forward conversion formulas above, but
    calculated for fp instead of lat.</li>
  <li>D = x/(N1k<sub>0</sub>)</li>
</ul>
<p>lat = fp - Q1(Q2 - Q3 + Q4), where:</p>
<ul>
  <li>Q1 = N1 tan(fp)/R1</li>
  <li>Q2 = (D<sup>2</sup>/2)</li>
  <li>Q3 = (5 + 3T1 + 10C1 - 4C1<sup>2</sup> -9e'<sup>2</sup>)D<sup>4</sup>/24</li>
  <li>Q4 = (61 + 90T1 + 298C1 +45T1<sup>2</sup>&nbsp; - 3C1<sup>2</sup> 
-252e'<sup>2</sup>)D<sup>6</sup>/720</li>
</ul>
<p>long = long0 + (Q5 - Q6 + Q7)/cos(fp), where:</p>
<ul>
  <li>Q5 = D</li>
  <li>Q6 = (1 + 2T1 + C1)D<sup>3</sup>/6</li>
  <li>Q7 = (5 - 2C1 + 28T1 - 3C1<sup>2</sup> + 8e'<sup>2</sup> + 24T1<sup>2</sup>)D<sup>5</sup>/120</li>
</ul>
<h3>What Do The Formulas Mean?</h3>
<p>As the sketch above shows, because of the poleward curve of 
parallels, the 
footprint latitude is always greater than the true latitude. Q1 is just a
 
scaling coefficient and is constant for any given fp. The tangent term 
basically means the closer to the 
pole you are, the faster the parallels curve. Q2 is a quadratic term in 
x. 
Again, as with converting from geographic coordinates to UTM, we 
approximate the 
parallel as a parabola and add higher order corrections.</p>
<p>To determine longitude, we make a simple minded approximation that 
longitude 
is proportional to easting, but then, since fp is too large, the true 
longitude 
is smaller, since it lies on a parallel closer to the the equator. The 
divisor 
cos(fp) corrects for the varying length of degrees of longitude as 
latitude 
varies.</p>
<h2><a name="Spreadsheet">A Spreadsheet Program</a></h2>
<p>Before linking to the program, note:</p>
<ul>
  <li>It is an Excel spreadsheet. If you have Excel on your computer, it
 will
    (or should) open when you click the link. Most major spreadsheet 
programs
    can read spreadsheets in other formats.</li>
	<li>A spreadsheet is not an applet or program. In particular, you can't
 
	manually enter data into a cell and preserve any formulas that are 
there. 
	That's why some aspects of data entry are clunkier than they might 
otherwise 
	be with, say, a Visual Basic program.</li>
  <li>There are three computation pages, one for single conversions, the
 other two
    for batch conversions. Other pages contain information on datums and
 the 
	specific conversion formulas. To use the batch conversions you need to 
be somewhat
    proficient in spreadsheets as you will have to copy data and cell 
formulas.</li>
  <li>For our mutual peace of mind, run a virus check.</li>
  <li>You may copy the program for your own non-commercial use and for
    non-commercial distribution to others, but not for commercial use. 
Please 
	give appropriate credit when citing your calculations. You may
    also modify it as needed for your personal use. In Internet 
Explorer, right 
	click on the link and select Save Target As... to save the spreadsheet 
to 
	your own disk.</li>
</ul>
<p><a href="http://www.uwgb.edu/dutchs/usefuldata/UTMConversions1.xls">Spreadsheet
 For UTM Conversion</a></p>
<h2><b>References</b></h2>
<p>Snyder, J. P., 1987; Map Projections - A Working Manual. U.S. 
Geological
Survey Professional Paper 1395, 383 p. <b>If you are at all serious 
about maps
you need this book.</b></p>
<p>Army, Department of, 1973; Universal Transverse Mercator Grid, U. S. 
Army
Technical Manual TM 5-241-8, 64 p.</p>

<p>NIMA Technical Report 8350.2, "Department of Defense World Geodetic
System 1984, Its Definition and Relationships with Local Geodetic 
Systems,"
Second Edition, 1 September 1991 and its supplements. The report is 
available
from the NIMA Combat Support Center and its stock number is 
DMATR83502WGS84.
Non-DoD requesters may obtain the report as a public sale item from the 
U.S.
Geological Survey, Box 25286, Denver Federal Center, Denver, Colorado 
80225 or
by phone at 1-800-USA-MAPS.</p>

<hr>



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